\[ \begin{aligned} \sigma_X^2&=E[(X-\mu_X)^2]\\ &=E[x^2-2\mu X+\mu_X^2]\\ &=E(X^2)-2\mu_XE(X)+\mu_X^2\\ &=E(X^2)-2\mu_X^2+\mu_X^2\\ &=E(X^2)-[E(X)^2] \end{aligned} \]
square distance between the data and expected value/center
where \(\mu_x=E(X)\) is the expected value of X.
An easier formula to use is
\[ \sigma_X^2=E(X^2)-[E(X)]^2 \]
\[ \sigma_X=\sqrt{\sigma_X^2} \]
Example: Let X be the random variable with PMF
| x | -1 | 1 | 2 |
|---|---|---|---|
| p(x) | 1/3 | 1/6 | 1/2 |
\[ E(X)=-1\times{1\over 3}+1\times{1\over 6}...={5\over 6} \]
Method 1:
\[ \begin{aligned} Var(X)=\sigma_X^2&=E[(X-E(x))^2]\\ &=(-1-{5\over 6})^2\times {1\over 3}+(1-{5\over 6})^2\times {1\over 6}+(2-{5\over 6})^2\times {1\over 2}\\ &=1.81 \end{aligned} \] Method 2:
\[ \begin{aligned} \sigma_X^2&=E(X^2)-[E(X)]^2\\ &=(-1)^2*{1\over 3}+(1)^2*{1\over 6}+(2)^2*{1\over 2}-({5\over 6})^2\\ &=2.5-({5\over 6})^2\\ &=1.81 \end{aligned} \]
For continuous variable:
Let \(X\sim U(-3,2)\). Find Var(X).
\[ E(X)=-{1\over 2}(middle~point)\\ f(x)={1\over 5},x\in[-3,2]\\ \]
\[ \begin{aligned} E(X^2)&=\int_{-\infty}^{\infty}x^2f(x)dx\\ &=\int_{-3}^{2}x^2{1\over 5}dx\\ &={1\over 15}(2^3-(-3)^2)\\ &={7\over3} \end{aligned} \]
\[ Var(X)={7\over 3}-(-{1\over 2})^2=2.08 \]
Method 2:
\[ Var(X)=\int_{-\infty}^{\infty}(x-\mu_X)^2f(x)dx=... \]
Transform of random variable
If the variance of X is \(\sigma_X^2\) and \(Y=a+bX\), then
\[ \sigma_Y^2=b^2\sigma_X^2 \]
Let X be a continuous random variable with CDF F and \(\alpha\) a number between 0 and 1. the 100(1-\(\alpha\))-th percentile of X is \(x_\alpha\) such that
\[ F(x_\alpha)=P(X\le x_\alpha)=1-\alpha \]
alpha as the upper area of the probability, eg. \(x_0.05\) is the area of top 0.05, as 95th percentile
Example: Let X be a continuous random variable with PDF
\[ f(x)\begin{cases}e^{-x},~~~x>0\\0,~~~ elsewhere\end{cases} \]
Find the IQR.
\[ F(Q_1)=0.25=0.288\\ 1-e^x=0.25\\ F(Q_3)=0.75=1.386\\ 1-e^x=0.75\\ IQR=Q_3-Q_1=1.098 \]
\[ M_x(t)=E(e^{tx}),~~~t\in(-h,h)for~some~h>0 \]
\[ M_x(0)=1\\ {dM_x(t)\over dt}=M'(0)=E({d\over dx}e_{tx})=E(xe^{tx})=E(x)\\ M^k(0)=E(x^k) \]
A Bernoulli trial or experiment is one whose outcome is either success or a faliure.
A Bernoulli random variable X takes the value of 1 if the outcome is a success and 0 if the outcome is a failure.
The probability of success is p.
Notation: \(X\sim Bern(p).\)
The PMF:
|x|0|1| |p(x)|1-p|p|